# 2.3: Subgroups

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Subgroups are non-empty subsets of a group that are themselves groups. In this section, we will explore how to define and characterize subgroups.

##### Definition: Subgroup

Let $$G$$ be a group. Let $$H$$ is a non empty subset of $$G$$ and $$(H,\ast)$$ is a group, then $$H$$ is a subgroup of $$G$$.  We write this as $$H \le G$$.

Trivial subgroups of $$G$$ are $$\{e\}$$ and $$G$$ (the group itself).

##### Note

To ensure a set is a group, we need to check for the following five conditions:

1. Non-empty set,

2. Closure,

3. Associativity,

4. Identity, and

5. inverse.

The following theorem allows us to check three conditions (rather than 5) to ensure a subset is a subgroup.

##### Theorem $$\PageIndex{1}$$

Let $$(G, \ast )$$ be a group then $$H \subseteq G$$ is a subgroup if and only if

The following conditions are satisfied:

1. The identity $$e \in G$$ is in $$H$$.

2. If $$h_1,h_2 \in H$$, then $$h_1 \ast h_2 \in H$$.

3. If $$h \in H$$ then $$h^{-1} \in H$$.

Proof

Assume that $$H \subseteq G$$.

We shall show that $$e_H=e_G$$, where $$e_H$$ is the identity of $$H$$ and $$e_H$$ is the identity of $$G$$.

Thus, it is clear that $$H \le G$$.◻

•
1. From (2), we have shown $$h_1 \ast h_2 \in H, \forall h_1,h_2 \in H$$.

2. From (1), we have shown the identity.

3. From (3), we have shown the inverses exist.

4. Every element of $$h_1,h_2,h_3 \in G$$ therefore associativity applies because $$H$$ inherits it from $$G$$ by definition.

•

Conversely, assume that $$H \subseteq G$$ satisfies 1, 2, and 3 shown in Theorem 1.

From (1), we have shown that $$H$$ is a non-empty set - it has at least an identity.

Since $$e_G$$ is the identity of $$G$$ and $$e_H \in G$$, $$e_Ge_H=e_He_G=e_H$$.

Further, $$e_H \in H, e_H^{-1}=e_H \in H$$.

Since $$e_H, e_H^{-1} \in G$$, $$e_He_H^{-1}=e_G$$, but $$e_He_H^{-1}=e_H$$.

Hence $$e_G=e_H$$.◻

Note: all subgroups of a group have a common element - the identity $$e$$. They could have more common elements. Since $$H$$ is a subset of $$G$$, $$e_H$$ is in $$H$$ by default.

2. We shall show that, if $$h_1,h_2 \in H$$, then $$h_1 \ast h_2 \in H$$.

Let $$h_1, h_2 \in H$$.

Since $$(H, \ast)$$ is a group, $$h_1 \ast h_2 \in H$$.◻

Note a group is closed by definition.

3. We shall show that,if $$h \in H$$ then $$h^{-1} \in H$$.

Since $$(H, \ast)$$ is a group, $$h^{-1} \in H$$.◻

Note each element in a group has an inverse by definition.

$$H \le G$$ iff:

$$H \ne \{\}$$.

$$(H, \ast)$$ is a group.  Need to check:

1. ,$$h_1 \ast h_2 \in H, \forall h_1,h_2 \in H$$

2. $$e \in H$$, where $$e \in G$$,

3. $$h^{-1} \in H, \forall \; h \in H$$,

4. $$h_1(h_2h_3)=(h_1h_2)h_3, \forall \; a,b,c \in H$$.

The following theorem allows us to check two conditions (rather than 5) to ensure a subset is a subgroup.

##### Theorem $$\PageIndex{2}$$

Let $$(G, \ast)$$ be a group.

Then $$H (\subseteq G)$$ is a subgroup of $$G$$ iff the following conditions are satisfied:

1.  $$e_G \in H$$, and

2.  $$g \ast h^{-1} \in H, \forall \; g,h \in H$$.

Proof:

Assume that $$H \le G$$.

Assume that $$H \subseteq G$$.

Since $$e \ast h_2 \in H, e \ast h_2^{-1} \in H$$.

Further, $$eh_2^{-1}=h_2^{-1} \in H$$.

Since $$h_1 \ast h_2 \in H, \ h_1 \ast (h_2^{-1})^{-1} \in H$$.

Therefore $$h_1 \ast h_2 \in H$$. ◻

It has an identity since $$H \ne \{\}$$.

See proof of (a) above

Associativity comes because $$h_1,h_2,h_3 \in H$$.

1. Let $$h_1, h_2 \in H$$.

Conversely, assume that $$H \subseteq G$$ satisfies (1) and (2) of Theorem 2.

We shall show $$H \le G$$.

$$H \ne \{\}$$.  Since we have a group, we have at least an identity.

$$h_1 \ast h_2 \in H, \forall h_1,h_2 \in H$$,

$$e \in H$$, where $$e \in G$$,

$$h^{-1} \in H, \forall \; h \in H$$,

$$h_1(h_2h_3)=(h_1h_2)h_3, \forall \; a,b,c \in H$$.

We shall show that $$g \ast h^{-1} \in H, \forall \; g,h \in H$$.

Then $$h^{-1} \in H$$ because $$H \subseteq G$$.

Since $$g,h^{-1} \in H$$, $$gh^{-1} \in H$$,

Hence the result.◻

$$H \le G$$ iff:

$$H \ne \{\}$$.

$$(H, \ast)$$ is a group.  Need to check:

We shall show that $$e_H=e_G$$, where $$e_H$$ is the identity of $$H$$ and $$e_H$$ is the identity of $$G$$.

Since $$e_G$$ is the identity of $$G$$ and $$e_H \in G$$, $$e_Ge_H=e_He_G=e_H$$.

Further, $$e_H \in H, e_H^{-1}=e_H \in H$$.

Since $$e_H, e_H^{-1} \in G$$, $$e_He_H^{-1}=e_G$$, but $$e_He_H^{-1}=e_H$$.

Hence $$e_G=e_H$$.◻

Let $$g,h \in H$$.

##### Example $$\PageIndex{1}$$

Consider $$(\mathbb{R}^*, \bullet)$$.  Decide if the following sets form subgroups of $$\mathbb{R}^*$$.

1. $$H=\{x\in \mathbb{R}^*|x=1$$ or $$x$$ is irrational $$\}$$.

No.

Counter example:

Consider $$\sqrt{2} \in H$$.

Further $$\sqrt{2} \bullet \sqrt{2}=2$$ but $$2 \not \in H$$

Therefore since there is no closure, $$H \not \le \mathbb{R}^*$$.◻

1. $$K = \{x \in \mathbb{R}^*| x \ge 1\}$$.

No.

Counter example:

Consider $$2 \in K$$.

Note that $$2^{-1}=\frac{1}{2}$$, but $$\frac{1}{2} \not \in K$$.

Thus the inverse of $$2$$ does not exist.

Since $$a^{-1} \not \exists \forall a \in K$$, $$K \not \le \mathbb{R}^*$$.◻

##### Example $$\PageIndex{2}$$

Let $$G = \Big( M_{22}(\mathbb{R), + \Big) }$$.

Suppose you take this matrix s.t. the trace of the matrix is zero.

Thus $$H=\Big\{ \begin{bmatrix} a & b\\ c & d \end{bmatrix} \Big| a+d=0 \Big\}$$.

Show that $$H \subseteq G$$.

Solution:

Note that $$\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$$ is the identity of $$G$$ and it is clearly an element of $$H$$.  Thus condition (1) of Theorem 2 is satisfied.

We will show that $$g \ast h^{-1} \in H, \forall \; g,h^{-1} \in H$$.

Let $$\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \begin{bmatrix} e & f\\ g & h \end{bmatrix} \in H$$.

Then $$a+d=0$$ and $$e+h=0$$ since the two matrices belong to $$H$$.

Consider  $$\begin{bmatrix} a & b\\ c & d \end{bmatrix} + \begin{bmatrix} -e & -f\\ -g & -h \end{bmatrix}=\begin{bmatrix} a-e & b-f\\ c-g & d-h \end{bmatrix}$$.

Then $$(a-e)+(d-h)=(a+d)-(e+h)=0-0=0$$.

Hence $$\begin{bmatrix} a-e & b-f\\ c-g & d-h \end{bmatrix} \in H$$.

Thus $$H \le G$$.

##### Definition: finite group

Let $$G$$ be a group.  If $$G$$ is finite and $$|G|=n$$ then we say that the order of $$G=n$$.

##### Theorem $$\PageIndex{3}$$

Subgroup test for finite groups:

Let $$H \subseteq G$$.

Then $$H \le G$$ iff:

1. $$H \ne \{\}$$. (non-empty).

2. $$a, b \in H$$ then $$ab \in H$$ (closed).

## Special subgroups

### Centralizer of  an element

##### Definition: Centralizer of an element

Let $$G$$ be a group.  For any $$a \in G$$, the Centralizer of $$a$$ in the group $$G$$,  $$C(a)=\{g \in G|ag=ga\}$$. Then  $$C(a)\le G$$.

##### Theorem $$\PageIndex{6}$$

Let $$G$$ be a group.  For any $$a \in G$$, the Centralizer of $$a$$ in the group $$G$$,  $$C(a)=\{g \in G|ag=ga\}$$. Then  $$C(a)\le G$$.

##### Example $$\PageIndex{1} Let \(G=GL(2,\mathbb{R})$$.  Then find

1. $$C \Bigg( \begin{bmatrix}1 & 1\\ 1 & 0 \end{bmatrix}\Bigg)$$.

2. $$C \Bigg( \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\Bigg)$$.

Solution:

Let $$\begin{bmatrix} a & b\\ c & d \end{bmatrix} \in G$$, Then $$a,b,c,d \in \mathbb{R}$$, and $$ad-bc \ne 0.$$

Consider $$C \Bigg( \begin{bmatrix}1 & 1\\ 1 & 0 \end{bmatrix}\Bigg)$$ means that $$\begin{bmatrix}1 & 1\\ 1 & 0 \end{bmatrix}\begin{bmatrix}a & b\\ c & d \end{bmatrix}=\begin{bmatrix}a & b\\ c & d \end{bmatrix}\begin{bmatrix}1 & 1\\ 1 & 0 \end{bmatrix}$$.

Thus $$\begin{bmatrix}a+c & b+d\\ a & b \end{bmatrix}=\begin{bmatrix}a+b & a\\ c+d & c \end{bmatrix}$$.

Thus $$a+b=a+c$$, $$a=b+d$$, $$c+d=a$$ and $$c=b$$.

Thus $$C\Bigg(\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}\Bigg) =\Bigg\{A= \begin{bmatrix} a & b\\ b & a-b \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det(A) \ne 0 \; \Bigg\}$$.

1. $$C \Bigg( \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\Bigg)$$.

Let $$\begin{bmatrix} a & b\\ c & d \end{bmatrix} \in G$$, Then $$a,b,c,d \in \mathbb{R}$$, and $$ad-bc \ne 0.$$

Consider $$C \Bigg( \begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix}\Bigg)$$ means that $$\begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix}a & b\\ c & d \end{bmatrix}=\begin{bmatrix}a & b\\ c & d \end{bmatrix}\begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix}$$.

Thus $$\begin{bmatrix}c & d\\ a & b \end{bmatrix}=\begin{bmatrix}b & a\\ d & c \end{bmatrix}$$.

Thus $$b=c$$ and $$d=a$$.

Thus $$C\Bigg(\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\Bigg) =\Bigg\{B= \begin{bmatrix} a & b\\ b & a \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det(B) \ne 0 \; \Bigg\}$$.

### Center of a group

##### Definition: Center

Let $$G$$ be a group. Then the center of $$G$$ is defined as,   $$Z(G)=\{x \in G| gx=xg \forall \; g \in G \}.$$

##### Theorem $$\PageIndex{3}$$

Let $$(G, \ast )$$ be a group then $$Z(G)$$ is a subgroup of $$G.$$

Proof:

Let $$G$$ be a group and $$g \in G$$.

We will show that $$Z(G)=\{x \in G| gx=xg \; \forall g \in G \}$$.

Let $$e$$ be the identity of $$G$$.

Since $$eg=ge \; \forall g \in G$$, $$e \in Z(G)$$.

Let $$x_1, x_2 \in Z(G)$$.

We shall show that $$x_1x_2^{-1} \in Z(G)$$.

Consider that $$x_2^{-1}g=gx_2^{-1}$$.

Further $$(x_1x_2^{-1})g=x_1gx_2^{-1}$$.

$$=(x_1g)x_2^{-1}$$

$$=(gx_1)x_2^{-1}$$

$$=g(x_1x_2^{-1})$$.◻

##### Note

Note that if $$G$$ is abelian, $$Z(G)=G$$.

##### Example $$\PageIndex{1}$$

Let $$G=GL(2,\mathbb{R})$$.  Then find $$\mathbb{Z}(G)$$.

Solution:

$$\mathbb{Z}(G)= \{ g \in G| gx=xg, \forall x \in G\}$$.

Let $$\begin{bmatrix} a & b c & d \end{bmatrix} \in G$$, Then $$a,b,c,d \in \mathbb{R}$$, and $$ad-bc \ne 0.$$

We have shown $$C\Bigg(\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}\Bigg) =\Bigg\{ \begin{bmatrix} a & b\\ b & a-b \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det \ne 0 \; \Bigg\}$$ and $$C\Bigg(\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\Bigg) =\Bigg\{ \begin{bmatrix} a & b b & a \end{bmatrix} \Bigg| a,b \in \mathbb{R}, \det \ne 0 \; \Bigg\}$$, $$\begin{bmatrix}1 & 1\\ 1 & 0 \end{bmatrix}, \begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix} \in G$$.

Thus for $$\mathbb{Z}(G)$$ to exists, $$\begin{bmatrix} a & b\\ b & a-b \end{bmatrix} = \begin{bmatrix} a & b\\ b & a \end{bmatrix}$$.

Thus since $$a-b=a$$, $$b=0$$.

Thus, $$\mathbb{Z}(G) =\Bigg\{ \begin{bmatrix} a & 0\\ 0 & a \end{bmatrix} \Bigg| a \in \mathbb{R}, a \ne 0 \; \Bigg\}$$.

##### Example $$\PageIndex{1}$$

Prove that for each element $$a \in G$$, where $$G$$ is a group, that $$C(a)=C(a^{-1})$$.

Proof:

Let $$x_1 \in C(a)$$ s.t. $$ax_1=x_1a$$.

We will show that $$x_1a^{-1}=a^{-1}x_1$$.

Consider that $$a^{-1}ax_1=a^{-1}x_1a$$

$$ex_1=a^{-1}x_1a$$

$$x_1=a^{-1}x_1a$$

$$x_1a^{-1}=a^{-1}x_1aa^{-1}$$

$$x_1a^{-1}= a^{-1}x_1e$$

$$x_1a^{-1}= a^{-1}x_1$$.

Thus, $$x_1 \in C(a^{-1})$$.

Hence $$C(a^{-1}) \subseteq C(a)$$.

Let $$x_2 \in C(a^{-1})$$ s.t. $$a^{-1}x_2=x_2a^{_1}$$.

Consider that $$aa^{-1}x_2=ax_2a^{-1}$$

$$ex_2=ax_2a^{-1}$$

$$x_2=ax_2a^{-1}$$

$$x_2a=ax_2a^{-1}a$$

$$x_2a= ax_2e$$

$$x_2a= ax_2$$.

Thus, $$x_2 \in C(a)$$.

Hence $$C(a) \subseteq C(a^{-1})$$.
Since $$C(a^{-1}) \subseteq C(a)$$ and $$C(a) \subseteq C(a^{-1})$$, $$C(a) = C(a^{-1})$$.◻

### Centralizer

##### Definition: Centralizer

Let $$G$$ be a group and $$H$$ be a subgroup of $$G$$ then  the Centralizer of $$H$$ in $$G$$ is  $$C(H)=\{g \in G|gh=hg \; \forall \;h \in H\}$$.

Note that if $$H=\{e\}$$ then $$C(H)=G$$.

##### Theorem $$\PageIndex{4}$$

Let $$G$$ be a group and $$H$$ be a subgroup of $$G$$ then  the Centralizer of $$H$$ in $$G$$ is  $$C(H)=\{g \in G|gh=hg \; \forall \;h \in H\}$$ is a subgroup of $$G.$$

Proof:

Let $$G$$ be a group and $$H$$ be a subgroup of $$G$$. We will show that $$C(H)$$ is a subgroup of $$G$$.

Let $$e$$ be the identity element of $$G$$, then $$e \in C(H)$$.

Let $$g_1,g_2 \in C(H)$$.

We will show that $$g_1g_2^{-1}h=hg_1g_2^{-1}$$.

Consider that $$g_1h=hg_1$$, $$g_2h=hg_2$$, $$\forall g \in H$$.

Since $$g_1h=hg_1$$

$$g_1hg_2^{-1}=hg_1g_2^{-1}$$

$$g_1g_2^{-1}h=hg_1g_2^{-1}$$.

Thus $$g_1g_2^{-1} \in C(H)$$

Hence $$C(H) \le G$$.

### Normalizer

Let $$G$$ be a group and $$H$$ be a subgroup of $$G$$.  For any $$x \in G$$, $$N(H)=xHx^{-1}=\{xhx^{-1}|h \in H\}$$.

This subgroup is called the Normalizer of $$H$$ in the group $$G$$.

##### Definition: Normalizer

Let $$G$$ be a group and $$H$$ be a subgroup of $$G$$. Then

for any $$x \in G, N_x(H)=xHx^{-1}=\{xhx^{-1}|h \in H\}$$, and

the Normalizer of $$H$$ in the group $$G$$, $$N(H)= \{ xHx^{-1} | x \in G\}.$$

##### Theorem $$\PageIndex{5}$$

Let $$G$$ be a group and $$H$$ be a subgroup of $$G$$. Then  for each $$g \in G$$, $$N_g(H)=gHg^{-1}$$ is a subgroup of $$G.$$

Proof:

Let $$g \in G.$$ Let $$e$$ be the identity of $$H$$ then $$geg^{-1} \in gHg^{-1}$$.

Consider $$geg^{-1}=(ge)g^{-1}=gg^{-1}=e$$.  Thus $$e \in gHg^{-1}$$.

Let $$a,b \in gHg^{-1}$$.

We will show that $$ab^{-1} \in H$$.

Consider $$a=gh_1g^{-1}$$ and $$b=gh_2g^{-1}$$, $$h_1,h_2 \in H$$.

Then $$ab^{-1}=(gh_1g^{-1})(gh_2g^{-1})^{-1}$$

$$=(gh_1g^{-1})((g^{-1})^{-1}h_2^{-1}g^{-1})$$

$$=gh_1g^{-1}gh_2^{-1}g^{-1}$$

$$=gh_1h_2^{-1}g^{-1} \in gHg^{-1}$$.

Thus since $$h_1, h_2^{-1} \in N_g(H)$$ and $$e \in N_g(H)$$, $$N_g(H)$$ is a subgroup of $$G$$.◻

This page titled 2.3: Subgroups is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.